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I saw a problem that read:

Show that if $A$ is an $n \times n$ integer matrix where $\det(A) = \pm 1$, then $A$ defines an invertible map on the torus $\mathbb{T}^n = \mathbb{R}^n / \mathbb{Z}^n$. That is, show that if $v - w \in \mathbb{Z}^n$, then $A(v) - A(w) \in \mathbb{Z}^n$ for all $v, w \in \mathbb{R}^n$. Show that this is otherwise not true.

Proving the bold if-then statement doesn't look too hard: we do after all have an integer matrix $A$, so $$ Av - Aw = A(v - w) = \begin{pmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nn} \end{pmatrix} \begin{pmatrix} v_1 - w_1 \\ \vdots \\ v_n - w_n \end{pmatrix} = \begin{pmatrix} \sum \limits_{j = 1}^{n} a_{1j} d_j \\ \vdots \\ \sum \limits_{j = 1}^{n} a_{nj} d_j \end{pmatrix}$$ where $d_j := v_j - w_j$.

Since all the $a_{ij}$'s are integers (given) and all the $d_j$'s are as well (because $v - w \in \mathbb{Z}^n$), then multiplication and addition of integers gives that $A(v - w)$ has integer entries.

My question: where does the $\det(A) = \pm 1$ come into play? How are the two statements (the bold one and the one preceding it) equivalent?

Thanks in advance.

JOlv
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1 Answers1

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The $\det(A) = \pm 1$ comes into play for the "invertible" part. You've shown that if $A$ is an integer matrix then $A$ induces a (not necessarily invertible) map on the torus. You want to show that you can find an inverse for $A$ on the torus if $\det(A) = \pm 1$.

For your second question, the bold statement is both equivalent to $A$ being an integer matrix and also $A$ inducing a well-defined map on $\mathbb{T}^n$ (i.e. these three statements are equivalent). Regarding the statement on well-defined maps, $A$ is a map on $\mathbb{R}^n$ and recall that you can try to define a map on equivalence classes by applying $A$ to equivalence class representatives. The induced map is well-defined if representatives of the same equivalence class are equivalent after applying $A$, i.e. if $\bar{w} = \bar{v}$ then $\overline{A(w)} = \overline{A(v)}$. In the case of $\mathbb{T}^n$, vectors $w, v \in \mathbb{R}^n$ are in the same equivalence class if $w - v \in \mathbb{Z}^n$ so the bold condition is equivalent to the induced map $A:\mathbb{T}^n \to \mathbb{T}^n$ being well-defined.

rso
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  • I'm starting to see, thank you for your answer. A follow-up question: why is $\det(A) = \pm 1$ necessary for getting an invertible map on $\mathbb{T}^n$? For general matrices, you only need $\det(A) \neq 0$, so why the specific values for the torus? Is it because torii are defined as surfaces of revolutions (with circles) and so any other value would mean getting a point no longer on the torus (i.e. outside or inside it, not on its surface)? – JOlv Oct 14 '24 at 05:41
  • @JOlv The specific value comes from the fact that if $A$ is an invertible integer matrix, then $A^{-1}$ is an integer matrix if and only if $\det(A) = \pm 1$. There's actually a SE question about it. You can then define $A^{-1}$ as a function on the torus, which is the inverse for $A$ on the torus. Off the top of my head, I am not sure if this condition is necessary for $A$ having an inverse on the torus, but it is definitely necessary if you are looking for an inverse induced by a matrix. – rso Oct 14 '24 at 05:53